Econometrics
Regression Problem Set
Due date: 2/8/10
Problems
1) Problem 4.2 in Stock and Watson (SW)
(Note: 1 lb = 2.54 in and 1 lb = .4536 kg)
2) Problem 4.5 in SW
3) Problem 4.6 in SW
4) Problem 4.7 in SW
5) Suppose that you run the regression:
Yi  0  1 X i  ui
Using ordinary least squares, you get the following estimates of the regression
coefficients:
ˆ1  
( X i  X )(Yi  Y )
 ( X i  X )2
and
ˆ0  Y  ˆ1 X
Now suppose you define a new variable Zi = 2Xi
and you run the regression:
Yi   0  1Zi  ei
Using ordinary least squares, you get the following estimates of the new regression
coefficients:
ˆ1  
( Zi  Z )(Yi  Y )
 (Z
2
i  Z)
and
ˆ 0  Y  ˆ1Z
Your job: Express ̂ 0 and ̂1 as a function of ̂ 0 and ̂1 .
6) Problem 5.3 in Stock and Watson (SW)
7) Problem 5.4 in SW
8) Problem 5.5 in SW
(Assume that the sample size is over 1000)
9) Problem 5.10 in SW
(Hint: Let n0 denote the number of observation with X  0 and n1 denote the
n
number of observations with X  1; note that  i 1 X i  n1 ; )
Also, write a sentence to explain why it makes intuitive sense that ˆ1  Y1  Y0 .
1. (a) Substituting Height  70, 65, and 74 inches into the equation, the predicted weights are
176.39, 156.69, and 192.15 pounds.
(b) Weight  394  Height  394  15  591.
(c) We have the following relations: 1 in  254 cm and 1 lb  04536 kg. Suppose the
regression equation in the centimeter-kilogram space is
Weight  ˆ 0  ˆ1 Height .
The coefficients are ˆ 0  9941 04536  45092 kg; ˆ1  394  024536
 07036 kg per
54
cm. The R2 is unit free, so it remains at R2  081 .
2.
(a) ui represents factors other than time that influence the student’s performance on the
exam including amount of time studying, aptitude for the material, and so forth. Some
students will have studied more than average, other less; some students will have higher
than average aptitude for the subject, others lower, and so forth.
(b) Because of random assignment ui is independent of Xi. Since ui represents deviations
from average E(ui)  0. Because u and X are independent E(ui|Xi)  E(ui)  0.
(c) (2) is satisfied if this year’s class is typical of other classes, that is, students in this year’s
class can be viewed as random draws from the population of students that enroll in the
class. (3) is satisfied because 0  Yi  100 and Xi can take on only two values (90 and
120).
(d) (i) 49  0.24  90  70.6; 49  0.24 120  77.8; 49  0.24 150  85.0
(ii) 0.24  10  2.4.
3.
Using E(ui |Xi )  0, we have
E(Yi |Xi )  E(0  1Xi  ui |Xi )  0  1E( Xi |Xi )  E(ui |Xi )  0  1Xi 
4.
The expectation of ˆ 0 is obtained by taking expectations of both sides of Equation (4.8):
 

1 n 
E ( ˆ0 )  E (Y  ˆ1 X )  E    0  1 X   ui   ˆ1 X 
n i1 
 

n
1
  0  E ( 1  ˆ1 ) X   E (ui |Xi )   0 
n i1
where the third equality in the above equation has used the facts that ̂1 is unbiased so
E(   ˆ )  0 and E(u |X )  0.
1
1
i
i
5.
ˆ1  
( Zi  Z )(Yi  Y )
 (Z
i
 Z)
2
=
 (2 X  2 X )(Y  Y )  2 ( X  X )(Y  Y )  1 ˆ
4 ( X  X )
2
 (2 X  2 X )
i
i
i
i
2
i
2
i
1
So:
1
2
ˆ 0  Y  ˆ1Z  Y  ˆ1 (2 X )  Y  ˆ1 (2 X )  Y  ˆ1 X  ˆ0
We find that ̂1 
1 ˆ
1 and ̂0  ˆ0 .
2
6.
The 99% confidence interval is 1.5  {3.94  2.58  0.31) or 4.71 lbs  WeightGain
 7.11 lbs.
7.
(a) 3.13  1.47  16  $20.39 per hour
(b) The wage is expected to increase from $14.51 to $17.45 or by $2.94 per hour.
(c) The increase in wages for college education is 1  4. Thus, the counselor’s
assertion is that 1  10/4  2.50. The t-statistic for this null hypothesis is
2.50
t  1.470.07
 14.71, which has a p-value of 0.00. Thus, the counselor’s assertion can
be rejected at the 1% significance level. A 95% confidence for 1  4 is 4  (1.47
 1.97  0.07) or $5.33  Gain  $6.43.
8.
(a) The estimated gain from being in a small class is 13.9 points. This is equal to
approximately 1/5 of the standard deviation in test scores, a moderate increase.
 5.56, which has a p-value of 0.00. Thus the null
(b) The t-statistic is t act  13.9
2.5
hypothesis is rejected at the 5% (and 1%) level.
(c) 13.9  2.58  2.5  13.9  6.45.
9. Let n0 denote the number of observation with X  0 and n1 denote the number of
observations with
n
n
X  1; note that  i 1 X i  n1 ; X  n1|n; n1  i 1 X iYi  Y1 ;
1

n
( X i  X ) i 1 X  nX  n1 
n
2
i 1
n
Y  n1 Y1 
n0
n
2
i
2
n12
n

 n1 1 
n1
n

n1n0
n
; n1Y1  n0Y0   i 1Yi , so that
Y0
From the least squares formula
ˆ1 

in1 ( Xi  X )(Yi  Y ) in1 Xi (Yi  Y ) in1 XiYi  Yn1
 n

in1 ( Xi  X )2
i1 ( Xi  X )2
n1n0 |n
n
n
n
n

(Y1  Y )   Y  1 Y1  0 Y0   Y1  Y0
n0
n0 
n
n 
n 
n n n0
n
Y0  Y0
and ˆ0  Y  ˆ1 X   0 Y0  1 Y1   (Y1  Y0 ) 1  1
n 
n
n
n
n